What is the integral from x to y of (t^2-1) ?

The integral from x to y of (t^2-1) is -y+x+(y^3-x^3)/3

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integral from x to y of (t^2-1)

\int_{x}^{y} (t^{2} - 1) d t

- y + x + \frac{y ^{3} - x ^{3}}{3}

Solution steps

\int_{x}^{y} t^{2} - 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{x}^{y} 1 d t + \int_{x}^{y} t^{2} d t

\int_{x}^{y} 1 d t = y - x

\int_{x}^{y} t^{2} d t = \frac{y ^{3} - x ^{3}}{3}

= - (y - x) + \frac{y ^{3} - x ^{3}}{3}

Simplify

= - y + x + \frac{y ^{3} - x ^{3}}{3}

f^{^{'}} (x) = x^{- 3} + π x^{2} - 2

f (2) = 10 - 3 x

x \to 2 + lim (2 x)

\int e^{x^{3}} (3 x^{2}) d x

\int cos (x) sin^{3} (x) d x

What is the integral from x to y of (t^2-1) ?
The integral from x to y of (t^2-1) is -y+x+(y^3-x^3)/3